In the Young’s double slit experiment, the intensity of light at a point on the screen where the path difference is K, ( being the wavelength of light used). The intensity at a point where the path difference is will be [2014]

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Solution

Q.
In the Young’s double slit experiment, the intensity of light at a point on the screen where the path difference $λ$ is K, ( $λ$ being the wavelength of light used). The intensity at a point where the path difference is $λ / 4$ will be [2014]

1 K

2 K/4

3 K/2

4 zero

Ans.
(3)

Intensity at any point on the screen is

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$

where $I_{0}$ is the intensity of either wave and $ϕ$ is the phase difference between two waves.

Phase difference, $ϕ = \frac{2 π}{λ} \times Path difference$

When path difference is $λ,$ then

$ϕ = \frac{2 π}{λ} \times λ = 2 π$

$∴$ $I = 4 I_{0} \cos^{2} (\frac{2 π}{2}) = 4 I_{0} \cos^{2} (π) = 4 I_{0} = K$ ...(i)

When path difference is $\frac{λ}{4}$ , then

$ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2}$ $∴$ $I = 4 I_{0} \cos^{2} (\frac{π}{4}) = 2 I_{0} = \frac{K}{2} [Using (i)]$

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